Solution: What happens here is that the first application of integration by parts results in another integral that is best approached by using integration by parts.

Solution: Again, we use integration by parts twice. The subscripts on the variables below tell us during which application of integration by parts those variables were used.

Definite Integral Examples

This is also a good example of how you can sometimes use integration by parts with one of the functions being trivial (i.e., ).

Solution: When we use integration by parts, we simply keep in mind that we're evaluating the original integral from 1 to 5. That means that we must evaluate BOTH terms (the and the integral) given by integration by parts. Also, it helps to remember here that .

More Complicated Examples

Solutions: We need to use integration by parts, plus a trig identity, plus substitution in order to compute this integral.

By using the trig identity and solving for we can write the above as

Using the substitution (and so ) this then becomes

Recursive Examples

Solution: In this case, neither or will become simpler when differentiated, and so at the start it doesn't seem like integration by parts can help. But, something interesting happens.

The main thing to notice here is that the integral we are trying to find is on both sides of the equation, and since the right hand side version is negative, we add it to both sides. We get

and dividing both sides by 2 gives

Solution: There are two ways to consider this problem. One is to use the identity

and substitution. The other is to use the Pythagorean identity

and integration by parts. The latter is done here.

Adding the integral of to both sides gives

Reduction Formula Example

Prove the formula

Solution: We did a problem a lot like this in class, with specific numbers. You also had a similar problem on homework. The main idea is to start with one function as and the other as . Proceed as follows.